Chris gave Brian a four-digit code to disarm Chris’s security system.

Total number of ways = 4 * 9 = 36 since CCCW can be arranged in 4 ways and W can be corrected in 9 ways.


IMO A

Hi,

How can 0 be in the thousands place because if it would be , then it will make 3-digit code.
I guess it should be 9*3 (because there are 9 numbers and since 0 cannot taken thousands place it will not be counted.)
Hence, for thousands place, there should be 8 numbers (excluding 0)

So, Total ways = 27 + 8 = 35.

Please tell me if I am wrong.

Hi aarushisingla ,

0-9 are 10 digits and the one W is not correct so we have 9 digits left
Since In Lock , 0 can be also put. Lock Password can be 0000 or 0900 or 0090 or 0000.

Hope U got my Point. Please Do PM , if You didn't get it.

Regards,
Rajat Chopra

Suppose that Brian typed in “0000”. From the information given to us, we understand that the correct code contains three digits of “0” and one non-zero digit.

If the first digit is incorrect, the correct code could be any one of 1000, 2000, …, 9000 (nine codes).

Similarly, if the second digit is wrong, the correct code could be any one of 0100, 0200, …, 0900 (nine codes). Following the same logic, we see that If the third or fourth digits are wrong, there will be 9 possibilities for the correct code in each case.

Thus, to guarantee that Brian identifies the correct code, he needs to try at most 9 + 9 + 9 + 9 = 36 codes.

Note: Technically, if he only needs to “identify” the correct code, he needs at most 35 attempts. If he does not get the correct code in the first 35 attempts, he can then be sure that the correct code is the only remaining code that he did not try.

Answer: A

ScottTargetTestPrep Bunuel Kinshook

Please one basic confusion

Only one digit W (0-9) will be correct and only 1 arrangement will be correct from following 4 arrangements right ?

WCCC , CWCC, CCWC, CCCW

So doesn't it mean that he will have to try 10 x 3 (3 wrong arrangements with full 10 digits) + 9 (1 correct arrangement with 1 correct digit) = 39 attempts ?

Lets say the correct code is 0009. You are given 0000. Start from the beginning (left) and work your way to the right.

You try 0000 initially (from the question) and that doesn't work. So you try nine more times: 1000, 2000, ...., 9000. Okay, they all didn't work, so try the 2nd digit now.

You don't need to try 0000 again since you did that already. You try 0100, 0200, 0300, 0400, ...., 0900. None of those work, so try the 3rd digit.

You don't need to try 0000 again since you did that already. You try 0010, 0020, 0030,....., 0090. None of those work so you try the 4th digit.

You don't need to try 0000 again since you did that already. You try 0001, 0002, 0003, and then on the last try (you're very unlucky!) 0009 works!

So 36 additional tries were needed.

Nicely explained. Thanks a bunch.

Your reasoning is almost correct, just notice that there are not 10 but 9 possibilities for W in each case. It is given that Brian tries a code which does not work. For each of the arrangements WCCC, CWCC, CCWC and CCCW; Brian does not need to consider one digit because of that one code he tried in the beginning. For instance, suppose Brian tried 0000 in the beginning and the actual code was 0009. For the arrangement WCCC, Brian only needs to test the values 1-9 for W. For the arrangement CWCC, again he needs to only test 1-9, etc.

Thus, if you decrease the number of options for W by 1 and do the same calculation again, you will get the correct answer of 9 + 9 + 9 + 9 = 36 more tries.

Let the number entered be 0001, which is a wrong code.

Worst case scenario is that the code is 0000, and he checks the digits from left to right and in ascending order.

The first will be 1001 to 9001.......9 ways.
Next will be 0101 for the second digit till 0901....9 ways
Similarly 9 ways for third digit, and last number checked will be 0091.
Next for the 4th digit, we will start from 0002 and go upto 0009......8 ways

If we are still not able to open it, the correct code has to be 0000, and we do not require to test it.
Thus, we require 35 MORE attempts to IDENTIFY the correct code.

The question is wrongly worded. It should have been: how many more attempts will Brian need to guarantee that he opens the security system.

Could someone tell whether this is an acceptable solution:
->There are 40 ways in which he can try a different code.
->4 of those ways would be repeated.
Final Answer: 40-4=36 ways.

In agreement.

When he enters the 35th attempt, at that point he would know that the next code he types in will work. Therefore, he shouldn’t have to type it in.

The correct code has been “identified.”

Be that as it may, the answer can not be more than 36, so A.

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